Photo AI

Two particles A and B have mass 0.4 kg and 0.3 kg respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2010 - Paper 1

Question icon

Question 8

Two-particles-A-and-B-have-mass-0.4-kg-and-0.3-kg-respectively-Edexcel-A-Level Maths Mechanics-Question 8-2010-Paper 1.png

Two particles A and B have mass 0.4 kg and 0.3 kg respectively. The particles are attached to the ends of a light inextensible string. The string passes over a small... show full transcript

Worked Solution & Example Answer:Two particles A and B have mass 0.4 kg and 0.3 kg respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2010 - Paper 1

Step 1

Find the tension in the string immediately after the particles are released.

96%

114 rated

Answer

To find the tension in the string (T), we can apply Newton's second law to both particles A and B.

For particle A (mass = 0.4 kg):

0.4gT=0.4a(1)0.4g - T = 0.4a\quad (1)

For particle B (mass = 0.3 kg):

T0.3g=0.3a(2)T - 0.3g = 0.3a\quad (2)

Adding equations (1) and (2): 0.4g0.3g=0.4a+0.3a0.4g - 0.3g = 0.4a + 0.3a

This simplifies to: 0.1g=0.7a0.1g = 0.7a

From this, we can find acceleration a: a=0.1g0.7a = \frac{0.1g}{0.7}

Now substituting into (1) to find T: T=0.4g0.40.1g0.7T = 0.4g - 0.4 \cdot \frac{0.1g}{0.7}

Substituting g = 9.8 m/s² gives: T3.36 N or 12g35T \approx 3.36 \text{ N} \text{ or } \frac{12g}{35}

Step 2

Find the acceleration of A immediately after the particles are released.

99%

104 rated

Answer

Using the value of acceleration from the previous calculations: a=0.1g0.7=1.4m/s2a = \frac{0.1g}{0.7} = 1.4 \, \text{m/s}^2

Step 3

Find the further time that elapses until B hits the floor.

96%

101 rated

Answer

Using the kinematic equation: v=u+atv = u + at Given that the initial velocity (u) is 0 and the time (t) is 0.5 s: v=0+(1.4)(0.5)=0.7m/sv = 0 + (1.4)(0.5) = 0.7 \, \text{m/s}

Using another kinematic equation for distance (s): s=ut+12at2s = ut + \frac{1}{2}at^2 Substituting values gives: s=0(0.5)+12(1.4)(0.5)2=0.175ms = 0(0.5) + \frac{1}{2}(1.4)(0.5)^2 = 0.175 \, \text{m}

Using the equation again with u = 1 m (the height above the floor): 1=0.175+12(9.8)t21 = 0.175 + \frac{1}{2}(9.8)t^2

i.e. 10.175=4.9t21 - 0.175 = 4.9t^2

Solving for t gives: t=0.8254.90.57s or 0.566st = \sqrt{\frac{0.825}{4.9}} \approx 0.57 \, \text{s} \text{ or } 0.566 \, \text{s}

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;